How do you solve $x^{2} - 2 x - 1 = x + 3$ $x^2-2x-1=x+3$ ?

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How do you solve $x^{2} - 2 x - 1 = x + 3$ $x^2-2x-1=x+3$ ?

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Answer 1 · 2018-04-13T05:15:13

$x = 4$ $x=4$
$x = - 1$ $x=-1$

Explanation:

First, get $0$ $0$ on one side.
$(x^{2} - 2 x - 1)$ $(x^2-2x-1)$ $- (x + 3)$ $-(x+3)$ = $x + 3$ $x+3$ $- (x + 3$ $-(x+3$ )

$x^{2} - 3 x - 4 = 0$ $x^2-3x-4=0$

Next, factor this equation.
$(x - 4) (x + 1) = 0$ $(x-4)(x+1)=0$
Therefore,
$(x - 4) = 0$ $(x-4)=0$
$x = 4$ $x=4$
or
$(x + 1) = 0$ $(x+1)=0$ ,
$x = - 1$ $x=-1$

You can verify this using FOIL:
= $x^{2} + x - 4 x - 4$ $x^2+x-4x-4$
= $(x^{2} - 3 x - 4)$ $(x^2-3x-4)$

Answer 2 · 2018-04-13T05:22:02

$x = - 1$ $x=-1$
$x = 4$ $x=4$

Explanation:

$x^{2} - 2 x - 1 = x + 3$ $x^2-2x-1=x+3$

$x^{2} - 2 x - 1 - x - 3 = 0$ $x^2-2x-1-x-3=0$

$x^{2} - 3 x - 4 = 0$ $x^2-3x-4=0$

$x^{2} - 4 x + x - 4 = 0$ $x^2-4x+x-4=0$

$x (x - 4) + 1 (x - 4) = 0$ $x(x-4)+1(x-4)=0$

$(x + 1) (x - 4) = 0$ $(x+1)(x-4)=0$

$x + 1 = 0$ $x+1=0$
$x = - 1$ $x=-1$

$x - 4 = 0$ $x-4=0$
$x = 4$ $x=4$

Answer 3 · 2018-04-13T05:22:45

$x = - 1, x = 4$ $x=-1,x=4$

Explanation:

Consider the given equation
$x^{2} - 2 x - 1 = x + 3$ $x^2-2x-1=x+3$

Subtract $x + 3$ $x+3$ from both sides (Or bring all terms to the left-hand side)

$x^{2} - 2 x - 1 - x - 3 = x + 3 - x - 3$ $x^2-2x-1-x-3=x+3-x-3$

Thus,
$x^{2} - 2 x - 1 - x - 3 = 0$ $x^2-2x-1-x-3=0$

Add like terms
$x^{2} - 3 x - 4 = 0$ $x^2-3x-4=0$

Split middle term so as to obtain $(- 4)$ $(-4)$ as product and $(- 3)$ $(-3)$ as sum
$x^{2} + x - 4 x - 4 = 0$ $x^2+x-4x-4=0$

Take common factors out from paired terms
$x (x + 1) - 4 (x + 1) = 0$ $x(x+1)-4(x+1)=0$

Then,
$(x + 1) (x - 4) = 0$ $(x+1)(x-4)=0$
$x + 1 = 0$ $x+1=0$ or $x - 4 = 0$ $x-4=0$
$x = - 1$ $x=-1$ or $x = 4$ $x=4$

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